Prove that if is a linear transformation from a vector space to a vector space , then for any subspace in , is a subspace in .

Let and , the base field for the vector spaces. It means by definition that . Since is a subspace it means . Therefore, since is a linear transformation it means which means . Thus, is closed under vector addition. Likewise, which means . Thus, is closed under scalar multiplications. All the other properties for being a vector space are satisfied because . Thus, is a vector space over .

Let and , the base field for the vector spaces. It means by definition that . Since is a subspace it means . Therefore, since is a linear transformation it means which means . Thus, is closed under vector addition. Likewise, which means . Thus, is closed under scalar multiplications. All the other properties for being a vector space are satisfied because . Thus, is a vector space over .

Thanks for your input; your proof confirms my belief that the same proof in my linear algebra textbook is incorrect; it messed up on x & y and f(x) & f(y) (and it wasted me a long time to try to make sense out of that flawed proof ).

PS: I think there's one part that your proof is incomplete; it hasn't showed that the zero vector is in

Using your logic, why must there be a condition that a subspace must contain a zero vector, since one of the other two conditions (i.e., closed under scalar multiplication) has already taken care of the first condition (by letting k=0 as you said)?

why must there be a condition that a subspace must contain a zero vector, since one of the other two conditions (i.e., closed under scalar multiplication) has already taken care of the first condition (by letting k=0 as you said)?

The condition is necessary, in order to exclude the possibility of the set being empty (the empty set is by convention not considered to be a subspace). If you're trying to show that X is a subspace then you can only make use of the implication if you know that there exists a vector x in X.